package com.jacklei.ch07;
/*
* 斐波那契数 （通常用 F(n) 表示）形成的序列称为 斐波那契数列 。该数列由 0 和 1 开始，后面的每一项数字都是前面两项数字的和。也就是：

F(0) = 0，F(1) = 1
F(n) = F(n - 1) + F(n - 2)，其中 n > 1
给定 n ，请计算 F(n) 。

 

示例 1：

输入：n = 2
输出：1
解释：F(2) = F(1) + F(0) = 1 + 0 = 1
示例 2：

输入：n = 3
输出：2
解释：F(3) = F(2) + F(1) = 1 + 1 = 2
示例 3：

输入：n = 4
输出：3
解释：F(4) = F(3) + F(2) = 2 + 1 = 3
 

提示：

0 <= n <= 30

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/fibonacci-number
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。*/
public class FibonacciNumber {
    public static void main(String[] args) {
        FibonacciNumber f = new FibonacciNumber();
        for (int i = 1; i < 50 ; i++) {
            System.out.print("n:\t" + i + "\t=>" + f.fib3(i));
            System.out.print("\tn:\t" + i + "\t=>" + f.fib2(i));
            System.out.print("\tn:\t" + i + "\t=>" + f.fib(i));
            System.out.println("\tn:\t" + i + "\t=>" + f.fib4(i));
        }
    }
    public int fib4(int n) {
        if (n < 2) {
            return n;
        }
        int[][] q = {{1, 1}, {1, 0}};
        int[][] res = pow4(q, n - 1);
        return res[0][0];
    }

    public int[][] pow4(int[][] a, int n) {
        int[][] ret = {{1, 0}, {0, 1}};
        while (n > 0) {
            if ((n & 1) == 1) {
                ret = multiply4(ret, a);
            }
            n >>= 1;
            a = multiply4(a, a);
        }
        return ret;
    }

    public int[][] multiply4(int[][] a, int[][] b) {
        int[][] c = new int[2][2];
        for (int i = 0; i < 2; i++) {
            for (int j = 0; j < 2; j++) {
                c[i][j] = a[i][0] * b[0][j] + a[i][1] * b[1][j];
            }
        }
        return c;
    }


    public int fib3 (int n){
        if(n == 0) return 0;
        if(n == 1) return 1;
        int[][] q = {{1,1},{1,0}};
        int [][] res = pow(q,n-1);
        return res[0][0];
    }

    private int[][] pow(int[][] q, int n) {
        int[][] ret = {{1,0},{0,1}};
        while(n >0){
            if((n & 1) == 1){
                ret = multiply(ret,q);
            }
            n >>= 1;
            q = multiply(q,q);
        }
        return ret;
    }

    private int[][] multiply(int[][] q, int[][] p) {
        int[][] c = new int[2][2];
        for (int i = 0; i < 2; i++){
            for ( int j = 0; j < 2; j++){
                c[i][j] = q[i][0] * p[0][j] + q[i][1] * p[1][j];
            }
        }
        return c;
    }

    public int fib2(int n){
        if(n == 0) return 0;
        if(n == 1) return 1;
        int f = 0 ;
        int s = 1;
        int cur = 0;
        for (int i = 2; i < n+1; i++) {
            cur = f + s;
            f = s;
            s = cur;
        }
        return cur;
    }

    public int fib1(int n) {
        int[] dp = new int[n];
        dp[0] = 0;
        dp[1] = 1;
        for (int i = 2; i < n; i++) {
            dp[i] = dp[i-1] + dp[i-2];
        }
        return dp[n -1] +dp[n -2];
    }
    public int fib(int n) {
        if(n == 0) return 0;
        if(n == 1) return 1;
        return fib(n-1) + fib(n-2);

    }

}
